what's the difference between those two regarding bit operations?

Question

say, a is an unsigned 32 bit integer.

a << 24
(a & 0xff)<<24

Similiar one, any difference for this two

a >> 16
(a >> 16) & 0xff

Answer 1

Your first example:

a << 24
(a & 0xff)<<24

In the case of a 32-bit unsigned integer, there is no difference in the result. From the C standard (6.5.7 Bitwise shift operators):

The result of E1 << E2 is E1 left-shifted E2 bit positions; vacated bits are filled with zeros. If E1 has an unsigned type, the value of the result is E1 × 2^E2, reduced modulo one more than the maximum value representable in the result type. (...)

Your second example

a >> 16
(a >> 16) & 0xff

The seconds line truncates the result to the range 0 .. 255, while the first does not. If you meant the difference between

a >> 16
(a >> 16) & 0xffff

then there is again no difference for unsigned 32-bit integers:

The result of E1 >> E2 is E1 right-shifted E2 bit positions. If E1 has an unsigned type or if E1 has a signed type and a nonnegative value, the value of the result is the integral part of the quotient of E1 / 2^E2. (...)

Note that "unsigned" is important here. For signed integers, the behaviour can be different, implementation-defined, or undefined.

Answer 2

Any difference for these two:

a >> 16
(a >> 16) & 0xff

Yes, you'll get different results for every a > 0x00FFFFFF.

Answer 3

a << 24
(a & 0xff)<<24

no difference between results, but the latter may avoid influencing overflow flag of register ?